Lecture 19 Practical Issues in PID Implementation Process Control - PowerPoint PPT Presentation

Lecture 19 Practical Issues in PID Implementation Process Control Prof. Kannan M. Moudgalya IIT Bombay Wednesday, 4 September 2013 1/27 Process Control Practical Issues in PID Implementation

Outline 1. Implementable derivative mode 2. Discretisation (minimal) 3. Handling communication mismatch 4. Setpoint kick 2/27 Process Control Practical Issues in PID Implementation

PID controller � t � e(t) + 1 de(t) � u(t) = u + K c e(w)dw + τ d τ i dt 0 3/27 Process Control Practical Issues in PID Implementation

1. Noise handling in derivative mode 4/27 Process Control Practical Issues in PID Implementation

Handling noise in error signal PID control law is given by � t e(t) + 1 de(t) � � ∆u(t) = K c e(w)dw + τ d τ i dt 0 ◮ Problem with derivative, thanks to noise ◮ Can be seen in Laplace transformed version also: ∆U(s) = K c (1 + 1 τ i s + τ d s)E(s) ◮ Can see that in Laplace transform also. function. Substitute s = j ω . To be explained in frequency response. 5/27 Process Control Practical Issues in PID Implementation

Implementable PID Controller 1 + 1 � � ◮ Recall ∆U(s) = K c τ i s + τ d s E(s) ◮ Problem happens at high s in the derivative mode: magnitude is unbounded ◮ Constrain its magnitude at high frequency ◮ Filtered derivative mode: � � 1 + 1 τ d s ◮ ∆u(t) = K c τ i s + e(t) 1 + τ d s N ◮ N is a large number ◮ Typical values are 5, 10, etc. 6/27 Process Control Practical Issues in PID Implementation

PID Control Implementation ◮ Discrete version is required for implementation ◮ Many ways to do this ◮ Digital Control by Moudgalya presents several methods ◮ We present one approach now 7/27 Process Control Practical Issues in PID Implementation

2. Introduction to Discretisation 8/27 Process Control Practical Issues in PID Implementation

PID Controller Discretisation Recall the PID control law: � t � e(t) + 1 de(t) � ∆u(t) = K c e(t)dt + τ d τ i dt 0 Let us discretise, with ˜ u(n) = ∆u(t) | t=n ˜ u(n) = K c [e(n)+ 1 � e(0) + e(1) e(n − 1) + e(n) � T s + · · · + T s τ i 2 2 e(n) − e(n − 1) � + τ D T s 9/27 Process Control Practical Issues in PID Implementation

Calculate Change in PID Control u(n) = K c [e(n)+ ˜ 1 � e(0) + e(1) e(n − 1) + e(n) � T s + · · · + T s τ i 2 2 e(n) − e(n − 1) � + τ D T s Calculate ˜ u(n) − ˜ u(n − 1) and simplify u(n) − ˜ ˜ u(n − 1) = s 0 e(n) + s 1 e(n − 1) + s 2 e(n − 2) 1 + T s + τ d τ d � � s 0 = K c , s 2 = K c 2 τ i T s T s − 1 + T s − 2 τ d � � s 1 = K c 2 τ i T s 10/27 Process Control Practical Issues in PID Implementation

3. Bumpless Control 11/27 Process Control Practical Issues in PID Implementation

Preferable form of control law ◮ Recall the discrete time control law (blue): ˜ u(n) − ˜ u(n − 1) = s 0 e(n) + s 1 e(n − 1) + s 2 e(n − 2) ◮ Should we calculate this as ∆˜ u(n) = f(e(n) , e(n − 1) , e(n − 2)) ◮ or as ˜ u(n) = g(˜ u(n − 1) , e(n) , e(n − 1) , e(n − 2)) ◮ What if ˜ u(n − 1) is not known exactly? 12/27 Process Control Practical Issues in PID Implementation

Communication errors: An example ◮ Consider manual to automatic mode change ◮ Suppose that the controller thinks that the valve position is 5% ◮ But the actual control valve position is at +50% position and that it is working fine ◮ Suppose that the control law gives ∆˜ u = 0 . 1% ◮ First control law will output this as ∆˜ u = 0 . 1% ◮ So, 50.1% will get implemented ◮ The second control law will output this as ˜ u(n) = ˜ u(n − 1) + 0 . 1% ◮ So, the second control law will enforce 5.1%! 13/27 Process Control Practical Issues in PID Implementation

Bumpless Control Because control law is given in terms of ∆˜ u, ◮ Mismatch between the state of end control element and what controller thinks it to be, may not matter much ◮ Hence known as bumpless transfer 14/27 Process Control Practical Issues in PID Implementation

4. Setpoint Kick 15/27 Process Control Practical Issues in PID Implementation

Recall PID Controller - Basic Design � t e(t) + 1 de(t) � � u(t) = K e(t)dt + τ d τ i dt 0 U(s) = K(1 + 1 τ i s + τ d s)E(s) = S c (s) △ U(s) R c (s)E(s) If integral mode is present, R c (0) = 0. Filtered derivative mode: � � 1 + 1 τ d s u(t) = K τ i s + e(t) 1 + τ d s N 16/27 Process Control Practical Issues in PID Implementation N is a large number, of the order of 10 to 100.

MCQ: Filtered derivative mode We use filtered derivative mode, because 1. We want to reduce the adverse effect of derivative mode 2. We want to make the controller stable 3. We want the offset to go to zero 4. We want the closed loop to be stable Answer: 1 17/27 Process Control Practical Issues in PID Implementation

Setpoint, Derivative, Proportional Kicks ◮ The standard PID control law has a shortcoming. ◮ If there is a sudden change in the setpoint, ◮ both proportional and derivative modes will introduce large jumps in the control effort, known as setpoint kick. ◮ The large change introduced by the derivative mode is known as derivative kick. ◮ Proportional mode could also produce a large control effort, known as proportional kick. ◮ Both derivative and proportional kicks are generally not acceptable. 18/27 Process Control Practical Issues in PID Implementation

MCQ: Setpoint Kick Setpoint kick implies 1. The setpoint becomes a large value 2. The control effort becomes a large value 3. Offset becomes a large value 4. The plant becomes unstable Answer: 2 19/27 Process Control Practical Issues in PID Implementation

Solution to Setpoint Kick Split the PID controller into two parts 20/27 Process Control Practical Issues in PID Implementation

2-DOF Controller y r T c u G = B R c A − S c R c u = T c r − S c y R c R c Arrive at the following relation between r - y. y = T c B / A BT c r = r R c 1 + BS c / AR c AR c + BS c Error e, given by r − y is given by BT c r = AR c + BS c − BT c � � e = 1 − r AR c + BS c AR c + BS c 21/27 Process Control Practical Issues in PID Implementation

Offset-Free Tracking of Steps with Integral Use capitals to denote Laplace transform: E(s) = A(s)R c (s) + B(s)S c (s) − B(s)T c (s) R(s) A(s)R c (s) + B(s)S c (s) s → 0 sA(s)R c (s) + B(s)S c (s) − B(s)T c (s) t →∞ e(t) = lim lim (A(s)R c (s) + B(s)S c (s))s Because the controller has integral action, R c (0) = 0: e( ∞ ) = S c (s) − T c (s) � = S c (0) − T c (0) � � S c (s) S c (0) � s=0 22/27 Process Control Practical Issues in PID Implementation

Previous slide - continued This can be satisfied if one of the following is met: T c = S c T c = S c (0) T c (0) = S c (0) 23/27 Process Control Practical Issues in PID Implementation

Solution to Setpoint Kick ◮ Split the PID controller ◮ Many solutions are possible - see Digital Control by Moudgalya 24/27 Process Control Practical Issues in PID Implementation

MCQ: Filtered derivative mode We use filtered derivative mode, because 1. We want to make the controller stable 2. We want the offset to go to zero 3. We want an implementable derivative control mode 4. We want the closed loop to be stable Answer: 3 25/27 Process Control Practical Issues in PID Implementation

What we learnt today ◮ Implementable derivative mode ◮ Discretisation ◮ Handling communication mismatch ◮ Setpoint kick 26/27 Process Control Practical Issues in PID Implementation

Thank you 27/27 Process Control Practical Issues in PID Implementation